Little help for my parabola?

• darkfire313
In summary, the conversation discusses an equation, f(x) = x^2 - 5x + 6, and its vertex. It is mentioned that the points 2 and 3 on the x axis reach 0, and the question is asked about the location of the vertex. The speaker provides two methods for finding the vertex, one using the fact that the vertex is halfway between two points of equal height and the other using the vertex form of a parabola. The conversation also briefly mentions calculus and the speaker asks the other person about their knowledge on the subject.

darkfire313

See, there is an equation which is f(x) = x^2 - 5x + 6. On the points 2 and 3 of the x axis, they reach 0. So where is the vertex? I'm guessing its -.25, but just to make sure i had to ask. Well, our class seemed to have a lot of it and our teacher, Ms. Knudsvig, didn't know.

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darkfire313 said:
See, there is an equation which is f(x) = x^2 - 5x + 6. On the points 2 and 3 of the x axis, they reach 0. So where is the vertex? I'm guessing its -.25, but just to make sure i had to ask. Well, our class seemed to have a lot of it and our teacher, Ms. Knudsvig, didn't know.
You mean that f(2)= 4- 10+ 6= 0 and f(3)= 9- 15+ 6= 0. Because this parabola has vertical axis, it vertex is exactly half way between 2 and 3, at 5/2. f(5/2)= 25/4- 25/2+ 6= 25/4- 50/4+ 24/4= -1/4= 0.25.

You could also get that, without using the fact that the vertex is exactly halfway between to points of equal height, by completing the square:

5/2= 5/2 and (5/2)^2= 25/4 so f(x)= x^2- 5x+ 25/4- 25/4+ 6= (x- 5/2)^2- 25/4+ 24/4= (x- 5/2)^2- 1/4.

Since a square is never negative, that is always -1/4 plus something. When x= 5/2, (5/2- 5/2)^2- 1/4= -1/4. The lowest point, the vertex, is (5/2, -1/4).

darkfire313 said:
See, there is an equation which is f(x) = x^2 - 5x + 6. On the points 2 and 3 of the x axis, they reach 0. So where is the vertex?

To find the vertex, all you have to do is write the equation in vertex form. Standard from for a parabola is ax^2+bx+c. Vertex form for a parabola is y= a(x-h)^2+k. Once you have the equation in this form, h and k are the coordinates for your vertex.

How much do you (darkfire) know about calculus?